What is arclength x=((y^4))/8+1/((4y^2)),1<= y<= 3 ?

The answer to arclength x=((y^4))/8+1/((4y^2)),1<= y<= 3 is 92/9

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Find the Arc Length x=(y^4)/8+1/(4y^2) , 1<=y<=3

arc length

x = \frac{( y ^{4} )}{8} + \frac{1}{( 4 y ^{2} )}, 1 \leq y \leq 3

\frac{92}{9}

Solution steps

Apply the arc length formula:

\int_{1}^{3} 1 + (\frac{y ^{3}}{2} - \frac{1}{2 y ^{3}})^{2} d y

Solve

\int_{1}^{3} 1 + (\frac{y ^{3}}{2} - \frac{1}{2 y ^{3}})^{2} d y : \frac{92}{9}

The arc length is:

= \frac{92}{9}

- \frac{d}{d x} (2 x)

- \frac{d}{d x} (4 e^{x})

- \frac{d}{d} y \frac{y}{x}

- \frac{d}{d x} y = sec (2 x)

\int (e^{a x}) d x

What is arclength x=((y^4))/8+1/((4y^2)),1<= y<= 3 ?
The answer to arclength x=((y^4))/8+1/((4y^2)),1<= y<= 3 is 92/9